Check if n is divisible by power of 2 without using arithmetic operators

Given two positive integers n and m, check whether n is divisible by 2^m without using arithmetic operators

Examples: 

Input: n = 8, m = 2
Output: Yes
Explanation: 2^2=4, and 8 is divisible by 4.

Input: n = 14, m = 3
Output: No
Explanation: 2^3=8, and 14 is not divisible by 8.

Using Bitwise Operators - O(1) Time and O(1) Space

The idea of this approach is based on the bitwise properties of powers of 2. A number is divisible by 2 if its least significant bit (LSB) is 0, divisible by 4 if the two least significant bits are 0, divisible by 8 if the three least significant bits are 0, and so on. Using this property, a number n is divisible by 2^m if its last m bits are all 0.

We can check this using the expression: n & ((1 << m) - 1)

Here’s how it works:

• 1 << m shifts the number 1 to the left by m positions, giving 2^m.
• (1 << m) - 1 creates a mask with the last m bits set to 1.
• n & ((1 << m) - 1) extracts the last m bits of n.
• If the result is 0, then n is divisible by 2^m; otherwise, it is not.

using namespace std;

// function to check whether n
// is divisible by pow(2, m)
bool isDivBy2PowerM(unsigned int n, unsigned int m)
{
    // if expression results to 0, then
    // n is divisible by pow(2, m)
    if ((n & ((1 << m) - 1)) == 0)
        return true;

    return false;
}

int main()
{
    unsigned int n = 8, m = 2;
    if (isDivBy2PowerM(n, m))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

class GFG {

    // function to check whether n
    // is divisible by pow(2, m)
    static boolean isDivBy2PowerM(int n, int m)
    {
        // if expression results to 0, then
        // n is divisible by pow(2, m)
        if ((n & ((1 << m) - 1)) == 0)
            return true;

        return false;
    }

    public static void main(String[] args)
    {
        int n = 8, m = 2;
        if (isDivBy2PowerM(n, m))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

# function to check whether n
# is divisible by pow(2, m)
def isDivBy2PowerM(n, m):

    # if expression results to 0, then
    # n is divisible by pow(2, m)
    if (n & ((1 << m) - 1)) == 0:
        return True

    return False

def main():
    n, m = 8, 2
    if isDivBy2PowerM(n, m):
        print("Yes")
    else:
        print("No")

if __name__ == "__main__":
    main()

using System;

class GFG {
    
    // function to check whether n
    // is divisible by pow(2, m)
    static bool isDivBy2PowerM(uint n, uint m)
    {
        // if expression results to 0, then
        // n is divisible by pow(2, m)
        if ((n & ((1U << (int)m) - 1)) == 0)
            return true;

        return false;
    }

    static void Main()
    {
        uint n = 8, m = 2;
        if (isDivBy2PowerM(n, m))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}

// function to check whether n
// is divisible by pow(2, m)
function isDivBy2PowerM(n, m)
{
    // if expression results to 0, then
    // n is divisible by pow(2, m)
    if ((n & ((1 << m) - 1)) === 0)
        return true;

    return false;
}

// driver code
let n = 8, m = 2;
if (isDivBy2PowerM(n, m))
    console.log("Yes");
else
    console.log("No");

Yes